1. ## [SOLVED] Function Question

Question:
The function $f : x \mapsto x^2 + 5x + 8$ is defined for the domain $x \leq a$, where a is a constant.
(i) Express $x^2 + 5x + 8$ in the form $(x + p)^2 + q$.
(ii) Find the smallest value of a for which $f$ has an inverse.
(iii) Find the domain of $f^{-1}$ corresponding to this value of $a$.

Attempt:
(i) $f(x) = x^2 + 5x + 8$

$= (x + \frac{5}{2})^2 - \left(\frac{5}{2}\right)^2 - 8$

$= (x + \frac{5}{2})^2 + \frac{7}{4}$

$p = \frac{5}{2} \mbox{ and }q = \frac{7}{4}$

Is the domain $= -\frac{5}{2}$ and range $= \frac{7}{4}$ ?

$= (x + \underbrace{\frac{5}{2}}_{Domain})^2 + \underbrace{\frac{7}{4}}_{Range}$

2. Draw the graph, easier to know domain and range,

And remember, domain of f(x)= range of finverse(x)
Range of f(x) = Domain of finverse(x)

3. Originally Posted by looi76
Question:
The function $f : x \mapsto x^2 + 5x + 8$ is defined for the domain $x \leq a$, where a is a constant.
(i) Express $x^2 + 5x + 8$ in the form $(x + p)^2 + q$.
(ii) Find the smallest value of a for which $f$ has an inverse.
(iii) Find the domain of $f^{-1}$ corresponding to this value of $a$.

Attempt:
(i) $f(x) = x^2 + 5x + 8$

$= (x + \frac{5}{2})^2 - \left(\frac{5}{2}\right)^2 - 8$

$= (x + \frac{5}{2})^2 + \frac{7}{4}$

$p = \frac{5}{2} \mbox{ and }q = {7}{4}$

Is the domain $= -\frac{5}{2}$ and range $= \frac{7}{4}$ ?

$= (x + \underbrace{\frac{5}{2}}_{Domain})^2 + \underbrace{\frac{7}{4}}_{Range}$
For the function $(x + \frac{5}{2})^2 + \frac{7}{4}$ the domain is $x \leq -\frac{5}{2}$ and the range is $y \geq \frac{7}{4}$.